The mean length of the first 20 space shuttle flights was about 7 days with a standard deviation of about 2 days. Using Chebychev's theorem, at least how many of the flights lasted between 3 and 11 days.

User Icon for Kuai Kuai answered
10 years ago

n = 20

µ = 7
σ = 2

(7-2(2), 7+ 2(2)

(µ-2σ, µ + 2σ)

k = 2

1-1/k^2

1- 1/(2)^2

1-1/4

(4-1)/4 = 3/4 = 0.75

at least (20)(.75) = 15 shuttle flights lasted between 3 days and 11 days.

User Icon for Clown Bot Clown Bot answered
1 year ago

Well, with Chebyshev's theorem, we can determine that at least 75% of the flights must fall within two standard deviations of the mean. Since the standard deviation is about 2 days, that means at least 75% of the flights will fall within the range of (7 - 2) to (7 + 2) days, which is 5 to 9 days.

Now, if we want to know how many flights lasted between 3 and 11 days, we can safely say that all of the flights within the range of 5 to 9 days will also fall within the range of 3 to 11 days. So, at least 75% of the flights will last between 3 and 11 days.

But hey, don't worry about those other 25% of flights that might be outside that range. Maybe they had some unexpected sightseeing stops along the way or got caught in interstellar traffic. Who knows? Space travel can be unpredictable sometimes!

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

Chebyshev's theorem states that for any dataset, regardless of its shape, at least (1 - 1/k^2) of the data falls within k standard deviations from the mean, where k is any positive constant greater than 1.

In this case, we can use Chebyshev's theorem to determine the minimum number of flights that lasted between 3 and 11 days.

First, let's calculate the number of standard deviations away from the mean that 3 days and 11 days are:

For 3 days:
(3 - 7) / 2 = -4 / 2 = -2

For 11 days:
(11 - 7) / 2 = 4 / 2 = 2

Now, let's calculate the fraction of data that falls within those bounds using Chebyshev's theorem:

For the lower bound (3 days):
At least 1 - 1/k^2 = 1 - 1/2^2 = 1 - 1/4 = 3/4 = 0.75

For the upper bound (11 days):
At least 1 - 1/k^2 = 1 - 1/2^2 = 1 - 1/4 = 3/4 = 0.75

Based on Chebyshev's theorem, at least 75% (3/4) of the flights lasted between 3 and 11 days.

To determine the minimum number of flights, we multiply the percentage by the total number of flights (20):
0.75 * 20 = 15

Therefore, at least 15 of the 20 space shuttle flights lasted between 3 and 11 days according to Chebyshev's theorem.

User Icon for Explain Bot Explain Bot answered
11 months ago

To solve this problem using Chebyshev's theorem, we need to find the range within k standard deviations from the mean, where k represents a value greater than 1.

Chebyshev's theorem states that for any set of observations, the proportion of data values within k standard deviations of the mean is at least 1 - 1/k^2, where k > 1.

In this case, we know that the mean length of the first 20 space shuttle flights is 7 days, with a standard deviation of 2 days. To find the range within 3 and 11 days, we need to calculate the number of standard deviations each value is from the mean.

First, let's calculate the number of standard deviations for each value:
- For 3 days: (3 - 7) / 2 = -2
- For 11 days: (11 - 7) / 2 = 2

Since we want to find the range between -2 and 2 standard deviations from the mean, where k > 1, we can use k = 2.

According to Chebyshev's theorem, at least 1 - 1/2^2 = 1 - 1/4 = 3/4 or 75% of the data values are within 2 standard deviations from the mean.

Using this information, we can conclude that at least 75% of the first 20 space shuttle flights lasted between 3 and 11 days.